)for a particular diamond mine, 77% of the diamonds fail to qualify as "gemstone grade". a random sample of 112 diamonds is analysed. find the probability that more than 81% of the sample diamonds fail to qualify as gemstone grade.

We use the z statistic:

z = (x – u) / s

But first let us calculate the standard deviation s, sample x and mean u.

s = sqrt (n p q)

s = sqrt (112 * 0.77 * (1 – 0.77))

s = 4.45

x = 0.81 * 112 = 90.72

u = 0.77 * 112 = 86.24

So the z score is:

z = (90.72 – 86.24) / 4.45

z = 1.00

From the standard tables, the P value at z = 1.00 using right tailed test is:

P = 0.1587 = 15.87%

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